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16 Apr 1998 (Updated )

Analysis and Solution of Discrete Ill-Posed Problems.

gravity(n,example,a,b,d)
function [A,b,x] = gravity(n,example,a,b,d)
%GRAVITY Test problem: 1-D gravity surveying model problem
%
% [A,b,x] = gravity(n,example,a,b,d)
%
% Discretization of a 1-D model problem in gravity surveying, in which
% a mass distribution f(t) is located at depth d, while the vertical
% component of the gravity field g(s) is measured at the surface.
%
% The resulting problem is a first-kind Fredholm integral equation
% with kernel
%    K(s,t) = d*(d^2 + (s-t)^2)^(-3/2) .
% The following three examples are implemented (example = 1 is default):
%    1: f(t) = sin(pi*t) + 0.5*sin(2*pi*t),
%    2: f(t) = piecewise linear function,
%    3: f(t) = piecewise constant function.
% The problem is discretized by means of the midpoint quadrature rule
% with n points, leading to the matrix A and the vector x.  Then the
% right-hand side is computed as b = A*x.
%
% The t integration interval is fixed to [0,1], while the s integration
% interval [a,b] can be specified by the user. The default interval is
% [0,1], leading to a symmetric Toeplitz matrix.
%
% The parameter d is the depth at which the magnetic deposit is located,
% and the default value is d = 0.25. The larger the d, the faster the
% decay of the singular values.

% Reference: G. M. Wing and J. D. Zahrt, "A Primer on Integral Equations
% of the First Kind", SIAM, Philadelphia, 1991; p. 17.

% Per Christian Hansen, IMM, November 18, 2001.

% Initialization.
if (nargin<2), example = 1; end
if (nargin<4), a = 0; b = 1; end
if (nargin<5), d = 0.25; end
if isempty(example), example = 1; end
if isempty(a), a = 0; end
if isempty(b), b = 1; end

% Set up abscissas and matrix.
dt = 1/n;
ds = (b-a)/n;
t = dt*((1:n)' - 0.5);
s = a + ds*((1:n)' - 0.5);
[T,S] = meshgrid(t,s);
A = dt*d*ones(n,n)./(d^2 + (S-T).^2).^(3/2);

% Set up solution vector and right-hand side.
nt = round(n/3);
nn = round(n*7/8);
x = ones(n,1);
switch example
case 1
   x = sin(pi*t) + 0.5*sin(2*pi*t);
case 2
   x(1:nt)    = (2/nt)*(1:nt)';
   x(nt+1:nn) = ((2*nn-nt) - (nt+1:nn)')/(nn-nt);
   x(nn+1:n)  = (n - (nn+1:n)')/(n-nn);
case 3
   x(1:nt) = 2*ones(nt,1);
otherwise
   error('Illegal value of example')
end
b = A*x;

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